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WEIRS
Classification of Weirs:
Design of Weirs:
Hydraulic Design
Structural Design
Floor Design
Detailed Drawings
Solved Example
Objectives of Weirs in Irrigation Canals
Proper distribution of water carried by a main canal among the branch canals depending
upon it
Reducing the hydraulic slope (gradient) in a canal (if canal water slope is greater than the
allowable water slope)
Reducing head on existing structures
Collecting sediments at US of structures (sand strap)
Weirs for reducing water slope in steep lands
Distance between weirs
ac = L * Slope (before)
ab = L * Slope (after)
rise (R) = ac – ab
= L {Slope (before) – slope (after)}
L = distance between weirs
L = R / (natural slope – required slope)
Classification of Weirs According to Geometrical Shape
Classification According to Position in Plan
Classification According to Dimensions of Cross Section
Classification According to Position of Down-Stream Water Level
a) Free- Overfall Weir (Clear-Overfall)
Q = 2/3 Cd B (2g) 0.5 H1.5



DSWL is lower than crest level
Q is independent of DSWL
QαH
b) Submerged Weir
Q = 2/3 Cd B (2g) 0.5 H1.5 + Cd B h1 (2gh2)0.5

DSWL is higher than weir crest
Q α H, h1, h2
Classification According to Crest Length (B)
Design of Weirs
Design of Weirs is divided to 3 parts:
I. Hydraulic Design (determination of crest level and weir length
according to head)
II. Structural Design (Empirical Dimensioning – check of stability)
III – Detailed Drawings
For proper Design of Water Structures:
Velocity of Flow:
Must cause minimum Loss in Head
Or
minimum Heading Up
Flow of Water in a Channel is controlled either by:


A Weir
or
A Regulator
Weirs:
Regulators:
For lands having steep slopes
For lands having mild slopes or flat lands
I- Hydraulic Design of Weirs
1- Clear Over fall Weir
Q = 2/3 Cd B (2g) 0.5 H 1.5
2 – Submerged Weir
Q = 2/3 Cd B (2g) 0.5 h21.5 + Cd B h1 (2*g*h2) 0.5
3 – Broad–Crested Weir
Q = 1.71 Cd B H 1.5
4 – Fayum Type Weir
Q = 1.65 B H 1.5
5 – Standing Wave Weir
Q = 2.05 B H 1.5
II Structural Design
1 The super structure
Theoretical Weir Profile
Scour Length of Weir Floor
Scour may be defined as deepening and widening of water channel under the influence of the flowing
water with high velocities.
The scour continues until the energy of the flowing water reaches the normal channel energy.
Velocity distribution through scour hole
Precautions against scour
Floor of Heading Up Structures
A weir on solid rock (impervious foundation) does not need long apron (Floor), but needs sufficient
width “b” to resist soil stresses.
A weir on pervious soil needs length “L” to:
a) Cover percolation length,
b) Resist scour from falling water
Definitions
Percolation is the flow of water under the ground surface due to an applied differential head
Percolation length (creep length) is the length to dissipate the total hydraulic pressure on the
structure
Undermining (Piping) is to carry away (wash) soil particles with flowing water below the
ground surface causing collapse or failure of the above structure
Determination of Percolation Length
To determine the critical head: (after which undermining occurs)
1- Measure Q for
different heads
2- H1 ----- Q1, v1= Q1 / A
H2 ------Q2, v2………. (k determined)
3- H……..Hn varies until Hcritical (soil particles begin to move)
Vcritical = Qcritical / A
vcr
vcr = k Hcr / L = K icr
k = vcr L / Hcr
L = K Hcr / vcr
= Qcr L / A Hcr
Soil
Clean gravel
Clean sand
Fine sand + silt
Clay
K (cm / min)
5000 – 50
50 – 0.05
0.05 – 0.00005
< 0.00005
Type of flow
Turbulent
Turbulent or laminar
Laminar
Always laminar
Permeability : (hydr. Conductivity)
Ability of fluid to move in the soil under certain head (dimensions of velocity)
v=ki
i=H/L
v α porosity + arrangement of grains
Seepage or percolation below weirs on previous soils:
-
a weir may be subject to failure from under seepage
-
water head will force (push) the water to percolate through the soil voids
-
if water velocity at D.S. end is not safe (> v critical) then undermining occurs, i.e. water at
exit will carry away soil particles
v = k I (Darcy,s law)
= k dP / dl = k H / L
In practice: icr = vcr / k
is unknown
Therefore we carry the 2nd experiment
e = voids ratio
e = vv/ vs
e = (1 – vs) / vs = (1 / vs) –1
Or 1+e = 1 / vs
or
vs = 1 / (1+e)
Upward force = H * A
Downward force = vs ( sat  1) * A * L
(net weight)
 SAT
= sp. Gr. Wt. Of soil under water
= (  SAT -1) A L / (1+e)
for stability: H. A. = (  SAT -1) A L / (1 + e)
H / L = icr = (  SAT - 1) / (1 + e)
Safe percolation length
can be determined
L = H / icr
Or L = H / icr (F.S.)
Values of icr & F.S.
Soil
Fine gravel
Coarse sand
Fine sand
Silt & clay
icr
0.25 – 0.20
0.20 – 0.17
0.17 – 0.14
0.14 – 0.12
F.S.
4 –5
5–6
6–7
7–8
If I > icr undermining (piping)
i.e. water has v >> to carry away soil particles
Bligh Creep Theory
The length of the seepage path transversed by the water is known as the length of creep (percolation
length).
Bligh supposed that the dissipation of head per unit length of creep is constant throughout the seepage
path.
CB = Bligh coefficient of percolation
Percolation length is the path length from (a) to (b)
LBligh = CBligh . H
L` = 2 t + L
C B = V/K
If L` > LB (Design is safe, no possibility of undermining)
If L` < LB (Design is unsafe, undermining occurs, leads to failure)
L` = L + 2 t + 2 S1 + 2 S2
L`  LB (design is safe, no possibility of undermining)
L` < LB (design is unsafe, undermining occurs, leads to failure)
Lane’s Weighted Creep Theory
Lane suggested that a weight of three should be given to vertical creep and a weight of one to horizontal
creep.
LL = CL H
Lane percolation length
L` = 1/3 L (horizontal) + L (vertical)
L` = 1/3 L + 2 t + 2 S1 + 2 S2
Distance between successive sheet piles
 d1 + d2

Distance between sheet piles a-a and b-b

Water percolation length takes the right path -----safe
Distance between sheet piles a-a & b-b < d1 + d2;
Water percolation length takes a short cut from a to b;
Actual percolation length is smaller than designed
unsafe
Design Head for Percolation
H = USHWL – DSHWL
(1)
H = USLWL – DSLWL
(2)
H = Crest level –DSBL
(3)
Design head H is the biggest of (1), (2), and (3)
Determination of Floor Dimensions
t1 = 0.5 – 1.0 m
t2 is taken 2.0 m
assumed
or
t2 = 0.8 (H)0.5
t3 = t 2 / 2  1 m
and l1 is assumed (1-2) H
L2 = is determined according to weir type  (3-8) m
LScour = Cs (Hs) 0.5
Or
LScour = 0.6 CB (Hs) 0.5
Hs = USHWL – DSBL – Yc
= Scour head; Yc = critical depth
Yc  1 / 3 q 2 / 2g
&
q=Q/B
where B is the weir length; q is the discharge per unit length
L` = l1 + l2 + ls + 2 t2
LB = CB . H
if L` LB no need for sheet pile
If L` < LB unsafe; use sheet pile
Depth of sheet piles = (LB – L`) / 2
Sheet pile depth  6 m
Determination of the uplift diagram

HD
h2 = H – t1/CB – l1 / CB
t2 = t / (γm) * Factor of safetyγ
t2 = F.S. [ h2 / (γm)] m.; γm = 2.2 t/m3
t2 = 1.3. [ h2 / (γm)]
then t3 = t2/2
≥
1 m.
t3 = F.S. [ h3 / (γm)] m
L3 = CB * h3 = x + t3
then the head h3 which corresponds to floor thickness t3
then get distance x
Precautions Against Percolation

The aprons are of plain concrete blocks of about 1.5 * 1 * 0.75 m deep

For small structure blocks of about 1 * 0.75 * 0.5 m deep may be used

The blocks are placed in rows with (70 – 100) mm open joints filled with broken stone.
An inverted filter of well graded gravel and sand is placed under the blocks in order to prevent
the loss of soil through the joints
EXAMPLE
A canal (A) is divided into two branches (i & ii).The discharge of branch (i)=2Q of branch (ii) at all
times. Two weirs have to be constructed at the entrance of each canal .
Data :- Bed width of canals (i & ii )
= ( 23.0 & 8.0 ) m .
- Flood discharge of canal (A)
= 105 cum/sec .
- Summer discharge of canal (A) =
45
cum/sec .
- DSHWL in the two canals = ( 11.00 )
- minimum water depth in the two canal branches = 4.0 m .
- Difference between H.W.L & L.W.L in canal(A) = .7 m .
- Submergence in canal (i)
= 1/3
- Bligh coeff. of percolation
= 16
- Bed level is constant in canal (A) and its branches .
- Q = 2 B H1.5
If a Board crested weir is constructed at the entrance of the two branches (i&ii) it is required to :-
1- Crest level of weirs ( i & ii ) .
2- Length of each weir .
3- HWL in canals (A) .
4- LWL in canal (A) & (i) .
5- Design of weir floor for canal (i) by applying Bligh method..
solution
QA = Qi + Qii
QA = 2 Qii + Qii
&
Qi =
2 Qii
At flood
QA = 105 = 3 Qii
Qii = 35 m3/s
& Qi = 70 m3/s
At summer
QA = 45 = 3 Qii
Qii = 15 m3/s
& Qi = 30 m3/s
For branch ( i )
Qmax /Qmin
= (2 B H11.5) / (2 B H21.5)
H1/H2 = (Qmax /Qmin )2/3
H1/H2 = 1.527 &
H1 - H2 =
.7
= H12/H22
= (70/30)2/3
H1 = 1.76 H2
(1)
(2)
From (1) & (2)
1.76 H2 - H2 = .7
H2 = .92 m
H1 = 1.62 m
h1/H1 = 1/3
h1 = 1.62/3
1- Crest level of weirs ( i & ii ) = 11 - .54 = ( 10.46 )
2- length of weir (i)
Qmax = 70 = 2 B (1.62)1.5
Qmin = 30 = 2 B (.92)1.5
B = 17 m
B = 17 m
B = 17 m
Length of weir (ii)
Qmax = 35 = 2 B (1.62)1.5
B = 8.5 m
Qmin = 15 = 2 B (.92)1.5
B = 8.5 m
B = 8.5 m
3- HWL in canals (A) = 10.46 + 1.62 = (12.08)
4- LWL in canal (A) = 10.46 + .92 = (11.38)
h2/H2 = 1/3
& h2 = .92/3 = .3
LWL in canal (i) = 10.46 + .3 = ( 10.76 )
Design of weir floor for canal (i) by applying Bligh method
BED LEVEL = 10.76 – 4 = 6.76
HD = 12.08 - 11
= 1.08
HD = 11.38 - 10.76 = .62
HD = 10.46 - 6.76
= 3.7
take
LB = CB * HD
HD = 3.7 m
= 16 * 3.7 = 59.2
Assume L1 = 6 m
L2 = 6 m
LS = CS (HS).5
CS = .6 CB
HS = 12.08 - 6.76 - Ycr
&
HS = 4.37
& LS = 20 m
Assume t2 = 2 m
L\ = 6 + 6 + 20 + 2 * 2 = 36
L\ < LB
unsafe use sheet pile d = (59.2 – 36) / 2 = 11.6
Use two sheet pile d =7 m & d = 5 m
h2 = 3.7 - .5/16 – 6/16- (2*7)/16 = 2.9
t2 = 2.9 * (1.3/1.2)
= 3.1 m
t3 = t2/2 = 1.6 m
> 1
1.6 = 1.3 * h3/1.2
L3 = 16 * 1.47 = X + 2*5 + 1.6
h3 = 1.47
&
X =
11.92 m